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how to get the most repeated element of a cell array?

4 Ansichten (letzte 30 Tage)
DuckDuck
DuckDuck am 26 Apr. 2013
Bearbeitet: Peter Saxon am 23 Jan. 2021
i have an cell array like this
{[];[];[];[];[];[];[];[];'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';}
is there a way to get the label of this array as rj?
  3 Kommentare
Matt J
Matt J am 26 Apr. 2013
maybe i did ask the question wrong, but if i have more {''} it will give me that name. instead i want {'rj'}. is there a workaround to counting?
Yes, you'll need to clarify the question. Why should {''} be ignored? Are there any other strings that should be ignored?
the cyclist
the cyclist am 26 Apr. 2013
In other words, you need to provide a more precise definition of the rule that defines the label.

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Akzeptierte Antwort

Cedric
Cedric am 26 Apr. 2013
Bearbeitet: Cedric am 26 Apr. 2013
the cyclist >> I knew I had overlooked something easier. :-)
Well, look at how I did overlook something easier ;-D :
C = {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'ab';'ab'} ;
setMatch = @(s,c) struct('string', s, 'count', c) ;
match = setMatch('', 0) ;
hashtable = java.util.Hashtable() ;
for k = 1 : length(C)
if isempty(C{k}), continue ; end
if hashtable.containsKey(C{k})
count = hashtable.get(C{k}) ;
if count >= match.count, match = setMatch(C{k}, count+1) ; end
hashtable.put(C{k}, count+1) ;
else
if match.count == 0, match = setMatch(C{k}, 1) ; end
hashtable.put(C{k}, 1) ;
end
end
Running this leads to;
>> match
match =
string: 'rj'
count: 3
  5 Kommentare
DuckDuck
DuckDuck am 26 Apr. 2013
i did it but i'm geting an error when trying to save results from columns with more than one string!!
for k=1:length(tsf)
cellData = labelw(:,tsf(k));
indexToEmpty = cellfun(@isempty,cellData);
cellData(indexToEmpty) = [];
[uniqueCellData(:,k),~,whichCell] = unique(cellData);
end;
Cedric
Cedric am 26 Apr. 2013
Bearbeitet: Cedric am 26 Apr. 2013
You don't need to save the content of temporary variables at each iteration of the loop. You just need to save results, and you should have something like (where the ".." have to be adapted to your case):
maxCountElements = cell(size(..), 1) ;
for k = 1 : ..
cellData = ..
indexToEmpty = cellfun(@isempty,cellData);
cellData(indexToEmpty) = [];
uniqueCellData = unique(cellData);
[uniqueCellData,~,whichCell] = unique(cellData);
cellCount = hist(whichCell,unique(whichCell));
[~,indexToMaxCellCount] = max(cellCount);
maxCountElements{k} = uniqueCellData(indexToMaxCellCount) ;
end

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Weitere Antworten (2)

the cyclist
the cyclist am 26 Apr. 2013
Bearbeitet: the cyclist am 26 Apr. 2013
Quite convoluted, but I think this works:
cellData = {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';}
indexToEmpty = cellfun(@isempty,cellData);
cellData(indexToEmpty) = {''};
uniqueCellData = unique(cellData);
[~,whichCell] = ismember(cellData,unique(uniqueCellData))
cellCount = hist(whichCell,unique(whichCell));
[~,indexToMaxCellCount] = max(cellCount);
maxCountElement = uniqueCellData(indexToMaxCellCount)
The essence of the algorithm is using the hist() function to count up the frequency. Unfortunately, that function only works on numeric arrays, so I had to use the ismember() command to map the cell array values to numeric values.
A further complication was the existence of the empty cell elements. I replaced them with empty strings. You'll need to be careful if your original array has empty strings.
  3 Kommentare
Matt J
Matt J am 26 Apr. 2013
No need for ismember,
[uniqueCellData,~,whichCell] = unique(cellData);
the cyclist
the cyclist am 26 Apr. 2013
I knew I had overlooked something easier. :-)

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Peter Saxon
Peter Saxon am 23 Jan. 2021
Bearbeitet: Peter Saxon am 23 Jan. 2021
Found a neat solution with categories, just posting this here so when I forget how to do this and google it again I'll see it...
C = {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'ab';'ab'} ;
catC=categorical(C);
catNames=categories(catC);
[~,ix] = max(countcats(catC));
disp(catName{ix}) % rj

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